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1850. Minimum Adjacent Swaps to Reach the Kth Smallest Number (Medium)

You are given a string num, representing a large integer, and an integer k.

We call some integer wonderful if it is a permutation of the digits in num and is greater in value than num. There can be many wonderful integers. However, we only care about the smallest-valued ones.

  • For example, when num = "5489355142": 
    <ul>
	<li>The 1<sup>st</sup> smallest wonderful integer is <code>"5489355214"</code>.</li>
	<li>The 2<sup>nd</sup> smallest wonderful integer is <code>"5489355241"</code>.</li>
	<li>The 3<sup>rd</sup> smallest wonderful integer is <code>"5489355412"</code>.</li>
	<li>The 4<sup>th</sup> smallest wonderful integer is <code>"5489355421"</code>.</li>
</ul>
</li>

Return the minimum number of adjacent digit swaps that needs to be applied to num to reach the kth smallest wonderful integer.

The tests are generated in such a way that kth smallest wonderful integer exists.

 

Example 1:

Input: num = "5489355142", k = 4
Output: 2
Explanation: The 4th smallest wonderful number is "5489355421". To get this number:
- Swap index 7 with index 8: "5489355142" -> "5489355412"
- Swap index 8 with index 9: "5489355412" -> "5489355421"

Example 2:

Input: num = "11112", k = 4
Output: 4
Explanation: The 4th smallest wonderful number is "21111". To get this number:
- Swap index 3 with index 4: "11112" -> "11121"
- Swap index 2 with index 3: "11121" -> "11211"
- Swap index 1 with index 2: "11211" -> "12111"
- Swap index 0 with index 1: "12111" -> "21111"

Example 3:

Input: num = "00123", k = 1
Output: 1
Explanation: The 1st smallest wonderful number is "00132". To get this number:
- Swap index 3 with index 4: "00123" -> "00132"

 

Constraints:

  • 2 <= num.length <= 1000
  • 1 <= k <= 1000
  • num only consists of digits.

Companies:
Google

Related Topics:
String, Greedy

Similar Questions:

Solution 1.

// OJ: https://leetcode.com/problems/minimum-adjacent-swaps-to-reach-the-kth-smallest-number/
// Author: github.com/lzl124631x
// Time: O(NK + N^2)
// Space: O(1)
class Solution {
public:
    int getMinSwaps(string s, int k) {
        auto start = s;
        for (int i = 0; i < k; ++i) {
            next_permutation(begin(s), end(s));
        }
        int ans = 0;
        for (int i = s.size() - 1; i >= 0; --i) {
            if (start[i] == s[i]) continue;
            int j = i - 1;
            while (j >= 0 && s[j] != start[i]) --j;
            ans += i - j;
            for (; j < i; ++j) s[j] = s[j + 1];
            s[i] = start[i];
        }
        return ans;
    }
};